Question

solve (t^(2)+25)(dx/dt)=(x^(2)+36)

Answer 1

Separate the variables to get
\[\begin{align*}
\frac{dx}{x^2+36}=\frac{dt}{t^2+25}
\end{align*}\]
For both integrals, you can make the appropriate trig sub, \((\text{ind var})=\sqrt{\text{constant}}~\tan(\text{new var})\).

Answer 2

In other words, as an example, set \(x=6\tan a\), then \(dx=6\sec^2a~da\), and the left side becomes
\[\int\frac{6\sec^2a}{36\tan^2a+36}~da\]

Answer 3

okay but i need to solve for x

Answer 4

Alright, so what's you're implicit solution?

Answer 5

Left hand side:
\[\begin{align*}\frac{1}{6}\int\frac{\sec^2a}{\tan^2a+1}~da&=\frac{1}{6}\int\frac{\sec^2a}{\sec^2a}~da\\\\
&=\frac{1}{6}\int da\\\\
&=\frac{1}{6}a\color{gray}{+C}\\\\
&=\frac{1}{6}\tan^{-1}\frac{x}{6}\color{gray}{+C}
\end{align*}\]
What's the right side going to be?

Answer 6

(1/5)arctan(x/5)+C

Answer 7

Well, x should be t, but yes, that's right.

Answer 8

So you have
\[\frac{1}{6}\tan^{-1}\frac{x}{6}=\frac{1}{5}\tan^{-1}\frac{t}{5}+C\]
Multiply both sides by 6:
\[\tan^{-1}\frac{x}{6}=\frac{6}{5}\tan^{-1}\frac{t}{5}+C\]
Take the tangent of both sides:
\[\frac{x}{6}=\tan\left(\frac{6}{5}\tan^{-1}\frac{t}{5}+C\right)\]
You can use an identity for the right side:
\[\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}\]

Answer 9

thank you!

Answer 10

Correct